Partial Solutions to Homework 08 - n have(HINT Remember...

MATH 74 HOMEWORK 8 1c Use part (a) and Lagrange’s theorem to show that for any positive integer n and any number a relatively prime to n , then a φ ( n ) ≡ 1 mod n . Solution. Since a is relatively prime to n , by problem 1 a ∈ ( Z /n Z ) × , so by Lagrange’s theorem, a raised to the order of G (which is φ ( n )) is the identity, so we conclude a φ ( n ) ≡ 1 mod n. 2 Prove Fermat’s theorem, that if p is a prime number and a is any integer, then a p ≡ a mod p . (HINT: φ ( p ) = p-1 and if a is less than p , a is relatively prime to p .) Solution. First consider the case where a < p so that a ∈ ( Z /p Z ) × . Then by Lagrange’s theorem, a p = a φ ( p )+1 = a φ ( p ) a 1 = a . Now if a > p , we may write a = b + rp where b < p . Then we compute: a ∼ = b ∼ = b p ∼ = b p-( rp ) p ∼ = a p mod p where the third equality comes the the binomial expansion of ( a-b ) p (every term but the ﬁrst and last is a multiple of p , so mod p we need only to consider the ﬁrst and last terms). 8 How many generators does a cyclic group of order

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Unformatted text preview: n have? (HINT: Remember Euler and his φ function.) Solution. We claim there are φ ( n ) generators. First, assume that G = Z /n Z which we may do since the number of generators will not change if we consider an isomorphic group. Now consider k ∈ Z /n Z . If k 6 = 1 divides n , then k added to itself n/k times will be n , so that k has order n/k . Since n/k 6 = n , k is not a generator. This show that the number of generators is at most φ ( n ). Conversely, assume that k is relatively prime to n . Then if k added to itself m times is n , we have km = n and so n must divide m , and hence k has order n . Thus k is a generator and we have that the number of generators is exactly φ ( n ). 1...
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